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How can i replace this?

How can I replace this
String str = "KMMH12DE1433";
String pattern = "^[a-z]{2}([0-9]{2})[a-z]{1,2}([0-9]{4})$";
String str2 = str.replaceAll(pattern, "repl");
Log.e("Founded_words2",str2);
What I got: KMMH12DE1433
What I want: MH12DE1433

Try it like this using a proper java.util.regex.Pattern and a java.util.regex.Matcher:
String str = "KMMH12DE1433";
//Make the pattern, case-insensitive using (?i)
Pattern pattern = Pattern.compile("(?i)[a-z]{2}([0-9]{2})[a-z]{1,2}([0-9]{4})");
//Create the Matcher
Matcher m = pattern.matcher(str);
//Check if we find anything
if(m.find()) {
//Use what you found - with proper capturing groups you
//gain access to parts of your pattern as needed
System.out.println("Found this: " + m.group());
}

If you just want to remove the first two characters and if the first two characters will always be uppercase letters:
String str = "KMMH12DE1433";
String pattern = "^[A-Z]{2}";
String str2 = str.replaceAll(pattern, "");
Log.e("Output string: ", str2);

try this :
String a = "KMMH12DE1433";
String pattern = "^[A-Z]{2}";
String rs = a.replaceAll(pattern,"");

Please change like this
String ans=str.substring(0);

How to find match for exact word using pattern matcher in java

I have shared my sample code here. here i am trying to find word "engine" with different strings. i used word boundary to match the words in string.
it matches word if it starts with #engine(example).
it should only match with exact word.
private void checkMatch() {
String source1 = "search engines has ";
String source2 = "search engine exact word";
String source3 = "enginecheck";
String source4 = "has hashtag #engine";
String key = "engine";
System.out.println(isContain(source1, key));
System.out.println(isContain(source2, key));
System.out.println(isContain(source3, key));
System.out.println(isContain(source4, key));
}
private boolean isContain(String source, String subItem) {
String pattern = "\\b" + subItem + "\\b";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(source);
return m.find();
}
**Expected output**
false
true
false
false
**actual output**
false
true
false
true

For this case, you have to use regex OR instead of word boundary. \\b matches between a word char and non-word char (vice-versa). So your regex should find a match in #engine since # is a non-word character.
private boolean isContain(String source, String subItem) {
String pattern = "(?m)(^|\\s)" + subItem + "(\\s|$)";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(source);
return m.find();
}
or
String pattern = "(?<!\\S)" + subItem + "(?!\\S)";

Change your pattern as below.
String pattern = "\\s" + subItem + "\\b";

If you are looking for a literal text enclosed with spaces or start/end of the string, you can split the string with a mere whitespace pattern like \s+ and check if any of the chunks equals the search text.
Java demo:
String s = "Can't start the #engine here, but this engine works";
String searchText = "engine";
boolean found = Arrays.stream(s.split("\\s+"))
.anyMatch(word -> word.equals(searchText));
System.out.println(found); // => true

Change the regexp to
String pattern = "\\s"+subItem + "\\s";
I'm using the
\s A whitespace character: [ \t\n\x0B\f\r]
For more info look into the java.util.regex.Pattern javadoc
Also if you want to support strings like these:
"has hashtag engine"
"engine"
You can improve it by adding the ending/starting line terminators (^ and $)
by using this pattern:
String pattern = "(^|\\s)"+subItem + "(\\s|$)";

Extract content after "=" and before "&", Regex expression in java

guys, I wanna extract the content in a string, the content is before "&" and after the "=", like this example:
asdfaf=afl10109&adsfjkl
I want to extract "afl10109" out of the string, can anyone teach me how to do this, I am very new to regex expression...

Use replaceAll() to replace the whole input with just what you want:
String target = str.replaceAll(".*=(.*)&.*", "$1");
The target is captured in a group (group number 1), which is then referenced in the replacement string.

try
public static void main(String args[]) {
String input="asdfaf=afl10109&adsfjkl";
Pattern pattern = Pattern.compile("=[^&]*&");
Matcher m = pattern.matcher(input);
while (m.find()) {
String str = m.group();
System.out.println( str.substring(1,str.length()-1));
}
}

This is not regex but you can also use split()
String str = "asdfaf=afl10109&adsfjkl";
System.out.println(str.split("=")[1].split("&")[0]);
Output:
afl10109

Using good old String#substring()
String str = "foo=bar&baz";
int begin = str.indexOf('=');
if (begin != -1) {
int end = str.indexOf('&', begin);
if (end != -1) {
System.out.println(str.substring(begin+1, end)); // bar
}
}

Replace different Regex-Matches with Match-based results in Java

One common usage for regex is the replacement of the matches with something that is based on the matches.
For example a commit-text with ticket numbers ABC-1234: some text (ABC-1234) has to be replaced with <ABC-1234>: some text (<ABC-1234>) (<> as example for some surroundings.)
This is very simple in Java
String message = "ABC-9913 - Bugfix: Some text. (ABC-9913)";
String finalMessage = message;
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
if (matcher.find()) {
String ticket = matcher.group();
finalMessage = finalMessage.replace(ticket, "<" + ticket + ">");
}
System.out.println(finalMessage);
results in<ABC-9913> - Bugfix: Some text. (<ABC-9913>).
But if there are different matches in the input String, this is different. I tried a slightly different code replacing if (matcher.find()) { with while (matcher.find()) {. The result is messed up with doubled replacements (<<ABC-9913>>).
How can I replace all matching values in an elegant way?

You can simply use replaceAll:
String input = "ABC-1234: some text (ABC-1234)";
System.out.println(input.replaceAll("ABC-\\d+", "<$0>"));
prints:
<ABC-1234>: some text (<ABC-1234>)
$0 is a reference to the matched string.
Java regex reference (see "Groups and capturing").

The problem is that the replace() method transforms the string over and over again.
A better way is to replace one match at a time. The matcher class has an appendReplacement-method for this.
String message = "ABC-9913, ABC-9915 - Bugfix: Some text. (ABC-9913,ABC-9915)";
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
String ticket = matcher.group();
matcher.appendReplacement(sb, "<" + ticket + ">");
}
matcher.appendTail(sb);
System.out.println(sb);

Splitting a string java

I have a string in format:
<+923451234567>: Hi here is the text.
Now I want to get the mobile number(without any non-alphanumeric characters) ie 923451234567 in the start of the string in-between < > symbols, and also the text ie Hi here is the text.
Now I can place a hardcoded logic, which I am currently doing.
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
String[] splitted = cpaMessage.getText().split(">: ", 2);
String mobileNumber=MyUtils.removeNonDigitCharacters(splitted[0]);
String text=splitted[1];
How can I neatly get the required strings from the string with regular expression? So that I don't have to change the code whenever the format of the string changes.

String stringReceivedInSms="<+923451234567>: Hi here is the text.";
Pattern pattern = Pattern.compile("<\\+?([0-9]+)>: (.*)");
Matcher matcher = pattern.matcher(stringReceivedInSms);
if(matcher.matches()) {
String phoneNumber = matcher.group(1);
String messageText = matcher.group(2);
}

Use a regex that matches the pattern - <\\+?(\\d+)>: (.*)
Use the Pattern and Matcher java classes to match the input string.
Pattern p = Pattern.compile("<\\+?(\\d+)>: (.*)");
Matcher m = p.matcher("<+923451234567>: Hi here is the text.");
if(m.matches())
{
System.out.println(m.group(1));
System.out.println(m.group(2));
}

You need to use regex, the following pattern will work:
^<\\+?(\\d++)>:\\s*+(.++)$
Here is how you would use it -
public static void main(String[] args) throws IOException {
final String s = "<+923451234567>: Hi here is the text.";
final Pattern pattern = Pattern.compile(""
+ "#start of line anchor\n"
+ "^\n"
+ "#literal <\n"
+ "<\n"
+ "#an optional +\n"
+ "\\+?\n"
+ "#match and grab at least one digit\n"
+ "(\\d++)\n"
+ "#literal >:\n"
+ ">:\n"
+ "#any amount of whitespace\n"
+ "\\s*+\n"
+ "#match and grap the rest of the string\n"
+ "(.++)\n"
+ "#end anchor\n"
+ "$", Pattern.COMMENTS);
final Matcher matcher = pattern.matcher(s);
if (matcher.matches()) {
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
}
I have added the Pattern.COMMENTS flag so the code will work with the comments embedded for future reference.
Output:
923451234567
Hi here is the text.

You can get your phone number by just doing :
stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">"))
So try this snippet:
public static void main(String[] args){
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
System.out.println(stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">")));
}
You don't need to split your String.